Why cadlag Processes have atmost countable discontinuities?

Question

Hi Guys I was trying to understand a proof on mathstack exchange by @Lord_Farin and I cannot really follow the proof. Can some body help me here?(Please see my doubts in bold)

Cardinality of set of discontinuities of cadlag functions

It clearly suffices to prove that there are only countably many discontinuities on any interval $[0,n]$.

The existence of left- and right-hand limits means that for all $x$ and for all $N$, there exists an $\epsilon_{x,N}$ with:

$$\forall y,z \in B(x; \epsilon_{x,N}): (y-x)(z-x) > 0 \implies |f(x)-f(z)| <\frac1N$$

where the antecedent ensures that $y$ and $z$ are on the same side of $x$.

Why is this last statement even true? . It would be clear if it was continuous but the author is just using existence of left and right limits. Where do we even use the point $y$

By compactness of $[0,n]$, finitely many of the $B(x;\epsilon_{x,N})$ cover $[0,n]$.

That is, except for finitely many points (the centers $x$ for the covering balls), we can be sure that no discontinuity of size bigger than $\frac1N$ exists in $[0,n]$.

This last statement is not clear to me at all. I mean the existence of finitely many open balls with center $x$ tells me if I assume (which i don't really understand why) what is written above is true then at these finitely many points there are no jumps/discontinuities?

Thus, for each $N$, the set:

$$\left\{x \in [0,n]: \left|\lim_{\xi\to x^+}f(\xi) - \lim_{\xi\to x^-}f(\xi)\right| > \frac1N\right\}$$

is finite. It follows that the set of discontinuities on $[0,n]$:

$$\left\{x \in [0,n]: \lim_{\xi\to x^+}f(\xi) \ne \lim_{\xi\to x^-}f(\xi)\right\} = \bigcup_{N \in\Bbb N}\left\{x \in [0,n]: \left|\lim_{\xi\to x^+}f(\xi) - \lim_{\xi\to x^-}f(\xi)\right| > \frac1N\right\}$$

is countable, and hence so is:

$$\left\{x \in [0,\infty): \lim_{\xi\to x^+}f(\xi) \ne \lim_{\xi\to x^-}f(\xi)\right\} = \bigcup_{n \in \Bbb N}\left\{x \in [0,n]: \lim_{\xi\to x^+}f(\xi) \ne \lim_{\xi\to x^-}f(\xi)\right\}$$

Answer 1

The first statement is false (and indeed, it is true if and only if $f$ is continuous at $x$), and is what confused you. The correct statement is:

$$\forall y,z \in B(x; \epsilon_{x,N}): (y-x)(z-x) > 0 \implies |f(y)-f(z)| <\frac1N.$$

Here's the rest of the argument.

For each $N$, $\cup_{x} B(x;\epsilon_{x,N})$ is an open cover of $[0,n]$ and therefore there exists a finite subcover. Let $\Lambda$ denote the set of centers of the subcovers, ranging over all $N\in {\mathbb N}$. Then $\Lambda$ is countable. Let $z \not \in \Lambda$. We show that $f$ is continuous at $z$. For every $N\in {\mathbb N}$ there exists $x_N \in \Lambda$ such that $z \in B(x_N;\epsilon_{x_N,N})$. Let $\delta_N=\min (|x_N-z|,\epsilon_{x_N,N}-|x_N-z|)$. Then if $|y-z|<\delta_N$, it follows that $y \in B(x_N;\epsilon_{x_N,N})$ as well as $(y-x_N)(z-x_N)>0$. As a result, $|f(y)-f(z)|<\frac{1}{N}$. Since $N$ is arbitrary, this implies that $f$ is continuous at $z$.